Equation of line thru (-1, -4), perpendicular to 3x - 2y + 9 = 0

mech649

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Find the equation of the line that contains the point (-1, -4) and is perpendicular to the line 3x - 2y + 9 = 0
 

stapel

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Find the equation of the line that contains the point (-1, -4) and is perpendicular to the line 3x - 2y + 9 = 0
What slope (here) did you obtain for the original line? For instance, did you solve for "y=" (here) and then read off the slope "m" from the form "y = mx + b"?

What slope, then, do you need to use for the perpendicular line?

Given this slope and the given points, what line equation (here) did you get?

Please be complete. Thank you! ;)
 

pka

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Find the equation of the line that contains the point (-1, -4) and is perpendicular to the line 3x - 2y + 9 = 0
Here is a useful result:
If \(\displaystyle Ax+By+C=0\) is a line then the line \(\displaystyle Bx-Ay+(Ay_0-Bx_0)=0\) is a line perpendicular to it through the point \(\displaystyle (x_0,y_0)\).
 

mech649

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Thank you!
 

mech649

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here is my work. My answer isnt the same as the answer in the book.
 

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stapel

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here is my work.

\(\displaystyle 3x\,−\,2y\,+\,9\,=\,0\)

\(\displaystyle \dfrac{−2y}{−2}\,=\, \dfrac{−3x}{−2}\,−\,\dfrac{9}{−2}\)

\(\displaystyle y\,=\,\dfrac{−3x}{−2}\,\dfrac{−9}{−2}\)
I think you mean the following:

. . . . .\(\displaystyle \,3x\,−\,3x\,−\,2y\,+\,9\,−\, 9\, =\,0\, −\, 3x\, −\, 9\)

. . . . .\(\displaystyle −2y\, =\, −3x\, −\, 9\)

. . . . .\(\displaystyle −2y\, =\, −3x\, +\, (−9)\)

. . . . .\(\displaystyle \dfrac{−2y}{−2}\, =\, \dfrac{−3x}{−2}\, +\, \dfrac{−9}{−2}\)

But why not make things easier for yourself in the first place?

. . . . .\(\displaystyle 3x\, −\, 2y\, +\, 2y\, +\, 9\, =\, 0\, +\, 2y\)

. . . . .\(\displaystyle 3x\, +\, 9\, =\, 2y\)

Then do the division, but by a positive value, so it's harder to lose signs (as you did in the last quoted line above):

. . . . .\(\displaystyle \dfrac{3x}{2}\, +\, \dfrac{9}{2}\, =\, \dfrac{2y}{2}\)

. . . . .\(\displaystyle \dfrac{3}{2}\, x\, +\, \dfrac{9}{2}\, =\, y\)

\(\displaystyle y\,=\,\dfrac{−3x}{−2}\,\dfrac{−9}{−2}\)

\(\displaystyle y\, =\, \dfrac{2}{3}\, x\, +\, b\)
How on earth did you get the second line above from the first? Why did the slope change from 3/2 to 2/3? Where did the y-intercept go?

My answer isnt the same as the answer in the book.
Well, yes; once the slope of the original line is lost, the negative reciprocal will be wrong, as must then be the line equation. Try starting again at the last good line (provided to you in corrected form below):

. . . . .\(\displaystyle y\, =\, \dfrac{3}{2}\, x\, +\, \dfrac{9}{2}\)

Please show all of your steps. Thank you! ;)
 
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